It is written in the Goldstein's Classical Mechanics text that $$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial {r_i}}{\partial {q_j}}\right) = \frac{\partial {\dot r_i}}{\partial {q_j}}=\sum_k \frac{\partial^2{r_i}}{\partial {q_j}\partial{q_k}}\dot q_k+\frac{\partial^2{r_i}}{\partial {q_j}\partial t}$$ where $$\dot r_i=\frac{\mathrm d}{\mathrm dt}r_i=\sum_k\frac{\partial{}r_i}{\partial{q_k}}\dot q_k+\frac{\partial {r_i}}{\partial t}$$ But it seems to me that there is another term in $\frac{\partial {\dot r_i}} {\partial {q_j}}$ because of product rule which is $$\sum_k \frac{\partial{r_i}}{\partial{q_k}}\frac{\partial{\dot q_k}}{\partial{q_j}}$$ which I think is equal to $$\frac{\partial{r_i}}{\partial{q_j}}\frac{\partial{\dot q_j}}{\partial{q_j}}$$ since $q_j$'s are independent among themselves.

Then how come $$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial {r_i}}{\partial {q_j}}\right) = \frac{\partial {\dot r_i}}{\partial {q_j}}?$$ Does $\frac{\partial{\dot q_j}}{\partial{q_j}} = 0\ $?

It is written in the Goldstein's Classical Mechanics text that $$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial {r_i}}{\partial {q_j}}\right) = \frac{\partial {\dot r_i}}{\partial {q_j}}=\sum_k \frac{\partial^2{r_i}}{\partial {q_j}\partial{q_k}}\dot q_k+\frac{\partial^2{r_i}}{\partial {q_j}\partial t},\tag{1.50b}$$ where $$\dot r_i=\frac{\mathrm d}{\mathrm dt}r_i=\sum_k\frac{\partial{}r_i}{\partial{q_k}}\dot q_k+\frac{\partial {r_i}}{\partial t}.\tag{1.46}$$ But it seems to me that there is another term in $\frac{\partial {\dot r_i}} {\partial {q_j}}$ because of product rule which is $$\sum_k \frac{\partial{r_i}}{\partial{q_k}}\frac{\partial{\dot q_k}}{\partial{q_j}},$$ which I think is equal to $$\frac{\partial{r_i}}{\partial{q_j}}\frac{\partial{\dot q_j}}{\partial{q_j}}$$ since $q_j$'s are independent among themselves.

Then how come $$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial {r_i}}{\partial {q_j}}\right) = \frac{\partial {\dot r_i}}{\partial {q_j}}~?\tag{1.50b}$$ Does $$\frac{\partial{\dot q_j}}{\partial{q_j}} = 0\ ?$$

It is written in the $\mathit\ Classical\ mechanics-\ Goldstein$ that $$\frac d{dt}\big(\frac{\partial {r_i}}{\partial {q_j}}\big) = \frac{\partial {\dot r_i}}{\partial {q_j}}=\sum_k \frac{\partial^2{r_i}}{\partial {q_j}\partial{q_k}}\dot q_k+\frac{\partial^2{r_i}}{\partial {q_j}\partial t}$$ where $$\dot r_i=\frac d{dt}r_i=\sum_k\frac{\partial{}r_i}{\partial{q_k}}\dot q_k+\frac{\partial {r_i}}{\partial t}$$ But it seems to me that there is another term in $\frac{\partial {\dot r_i}} {\partial {q_j}}$ because of product rule which is  $$\sum_k \frac{\partial{r_i}}{\partial{q_k}}\frac{\partial{\dot q_k}}{\partial{q_j}}$$ which I think is equal to  $$\frac{\partial{r_i}}{\partial{q_j}}\frac{\partial{\dot q_j}}{\partial{q_j}}$$ since $q_j$'s are independent among themselves.

Then how come $$\frac d{dt}\big(\frac{\partial {r_i}}{\partial {q_j}}\big) = \frac{\partial {\dot r_i}}{\partial {q_j}}$$ Does $\frac{\partial{\dot q_j}}{\partial{q_j}} = 0\ $?

It is written in the Goldstein's Classical Mechanics text that $$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial {r_i}}{\partial {q_j}}\right) = \frac{\partial {\dot r_i}}{\partial {q_j}}=\sum_k \frac{\partial^2{r_i}}{\partial {q_j}\partial{q_k}}\dot q_k+\frac{\partial^2{r_i}}{\partial {q_j}\partial t}$$ where $$\dot r_i=\frac{\mathrm d}{\mathrm dt}r_i=\sum_k\frac{\partial{}r_i}{\partial{q_k}}\dot q_k+\frac{\partial {r_i}}{\partial t}$$ But it seems to me that there is another term in $\frac{\partial {\dot r_i}} {\partial {q_j}}$ because of product rule which is  $$\sum_k \frac{\partial{r_i}}{\partial{q_k}}\frac{\partial{\dot q_k}}{\partial{q_j}}$$ which I think is equal to  $$\frac{\partial{r_i}}{\partial{q_j}}\frac{\partial{\dot q_j}}{\partial{q_j}}$$ since $q_j$'s are independent among themselves.

Then how come $$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial {r_i}}{\partial {q_j}}\right) = \frac{\partial {\dot r_i}}{\partial {q_j}}?$$ Does $\frac{\partial{\dot q_j}}{\partial{q_j}} = 0\ $?